How do you solve the quadratic equation by completing the square: #x^2 + 4x - 21 = 0#?

1 Answer
Jul 28, 2015

Answer:

#x_1 = -7#, #x_2 = 3#

Explanation:

Your starting quadratic equation looks like this

#x^2 + color(blue)(4)x - 21 = 0#

To solve this quadratic by completing the square, you need to add a term to both sides of the equation, such that the left side of the equation becomes equivalent to the square of a binomial.

First, move the term that does not contain #x# or #x^2# to the right side of the equation by adding #21# to both sides

#x^2 + 4x - color(red)(cancel(color(black)(21))) + color(red)(cancel(color(black)(21))) = 0 + 21#

#x^2 + 4x = 21#

To do that, you need to divide the coefficient of the #x#-term by 2, square it, then add the result to both sides of the equation.

In your case, you have

#color(blue)(4)/2 = 2#, then #2^2 = 4#

Your quadratic will become

#x^2 + 4x + 4 = 21 + 4#

The left side of the equation can be wrtitten as

#x^2 + 4x + 4 = (x+2)^2#

This means that you have

#(x+2)^2 = 25#

To solve this equation, take the square root from both sides of the equation to get

#sqrt((x+2)^2) = sqrt(25)#

#x + 2 = +-5 => x_(1,2) = +- 5 - 2#

The two solutions for this equation are

#x_1 = -5 -2 = color(green)(-7)#

#x_2 = +5 - 2 = color(green)(3)#